The square root of integer x is equal to the sum of the cubes of y and

We can create the following equation:

√x = y^3 + z^3

Since the squares of y and z are each less than 10, y < √10 and z < √10, which implies y^3 < 10√10 and z^3 < 10√10.

Thus:

√x < 10√10 +10√10

√x < 20√10

x < 4000

Since x is an integer, the greatest value of x is 3999.

Answer: D

Beats me why I should "root" the y instead of assuming a possible highest value for y in \(y^2\) e.g., y= 3 ?

Y and z need not be integers

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Hi,

Sorry for the colour. Didn't realise it.

You are WRONG when you read it as an integer.
It is nowhere given as an integer.
So if √10=3.16227766.., the value could be 3.16227765..
Hence we find answer by taking it as 10 and finding one value lesser INTEGER.

Hope it helps

perfect squares less than 10 are 1 ,4,9 for which value of y,z could be 1 ,2 ,3

but for greatest possible value of x we will take y= z= 3

now √x = y^3+z^3 =3^3 + 3^3

√x = 27+27 = 54

x= 54^2
x=2916

Answer A

Perfect squares less than 10 are 1 ,4,9 for which value of y,z could be 1 ,2 ,3
but for greatest possible value of x we will take y= z= 3
now √x = y^3+z^3 =3^3 + 3^3
√x = 27+27 = 54
x= 54^2
x=2916
Option A.

Problem states square root of integer x is equal to the sum of the cubes of y and z, hence to achieve that we cannot have y & z to be non integer values as we frame the equation x=y^3+z^3=10√10+10√10, as √10 when simplified would give a decimal component as well. Or in other words y & z cannot be √10 each.
Please advise.

Hi FB2017

Although I did the same as you did, but I was wrong. below is explanation-

we are not saying x=10√10+10√10 rather it is x<10√10+10√10 i.e x<20√10 (Note: if we take x=20√10 then we will get x=4000)

as x<20√10 we are saying that x<4000 i.e x=3999 as x should be an integer.


Kudo please if this helps you.

\(\sqrt{x} = y^3 + z^3\)
\(y^2 <10 , z^2 <10\)
\(-> y <\sqrt{10}, z <\sqrt{10}\)


So, \(\sqrt{x} < \sqrt{10}^3 + \sqrt{10}^3\)
\(\sqrt{x} < 20*\sqrt{10}\)
\(x < (20*\sqrt{10})^2\)
\(x<4000\)
\(So x = 3999\)

Answer D

Ans is clearly D

consider max values
y^2=10 means y^6 =1000 and same for z^6=1000. or y=z
x^0.5 = y^3+z^3
square both sides
x= y^6+z^6+2(yz)^3
= 1000+1000+2(z^2)^3
= 1000=1000+2(z^6)
=1000+1000+2(1000)
= 4000
but y and z are less than 10
therefore x is less than 4000 and integer value less than 4000 is 3999.


Give kudos take kudos

(1st)

Sum of Cubes:

(A)^3 + (B)^3 = (A + B) (A^2 - AB + B^2)


(2nd) if we are given that the square root of X equals the SUM of Y cubed and Z cubed, since we can not take the square root of a negative value, both Y and Z must be positive values


(3rd)

Given that

(Y)^2 < 10

(Z)^2 < 10


We can take the square root of both sides of the inequality without worrying about the negative value because Z and Y must be positive

0 < Y < sqrt(10)

0 < Z < sqrt(10)

(Lastly)

Assume that Y and Z are both equal to the value of their boundary ——-> sqrt(10)

Plugging in sqrt(10) into the Sum of Cubes formula given above we get

sqrt(X) = (2 * sqrt(10)) * (10 - sqrt(100) + 10)

And

sqrt(X) = (2 * sqrt(10)) * (20 - 10)

And

sqrt(X) = 20 * sqrt(10)


——squaring both sides to eliminate the square root we get——


X = (400) * (10)

Or

X = 4000

However, the values of Y and Z were LESS than the square root of 10, which we plugged into the Sum of Cubes formula.

Thus, the value for X must be less than this amount


X < 4000


D is the largest answer below

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